What is the solution for y^{''}-y^'-30y=0,y(0)=1,y^'(0)=5 ?

The solution for y^{''}-y^'-30y=0,y(0)=1,y^'(0)=5 is y=(1-1/11)e^{6t}+1/11 e^{-5t}

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y^{''}-y^'-30y=0,y(0)=1,y^'(0)=5

y^{^{''}} - y^{^{'}} - 30 y = 0, y (0) = 1, y^{^{'}} (0) = 5

y = (1 - \frac{1}{11}) e^{6 t} + \frac{1}{11} e^{- 5 t}

Solution steps

y^{''} (t) - y^{'} (t) - 30 y = 0

Solve linear ODE:

y = (1 - \frac{1}{11}) e^{6 t} + \frac{1}{11} e^{- 5 t}

y = (1 - \frac{1}{11}) e^{6 t} + \frac{1}{11} e^{- 5 t}

x^{^{''}} + k x = 0, x (0) = \frac{2}{3}, x^{^{'}} (0) = 5

y^{^{''}} - 2 y^{^{'}} - 143 y = 0, y (0) = 12, y^{^{'}} (0) = - 12

2 y^{^{''}} + 7 y^{^{'}} + 3 y = 0, y^{^{'}} (0) = 0, y (0) = 3

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = e^{3} x cos (2 x)

y^{^{''}} = \frac{( a - 2 )}{a} y

What is the solution for y^{''}-y^'-30y=0,y(0)=1,y^'(0)=5 ?
The solution for y^{''}-y^'-30y=0,y(0)=1,y^'(0)=5 is y=(1-1/11)e^{6t}+1/11 e^{-5t}